2y"+y'-y=x+1.
Let y=yc+yp, where yc is the characteristic solution and yp is the particular solution.
2yc"+yc'-yc=0 can be solved by factorising 2r2+r-1=(2r-1)(r+1) which leads to the characteristic solution:
yc=Ae½x+Be-x, where A and B are constants which can be found through applying the initial conditions.
For yp we "guess" that yp=ax+b where a and b are constants found by applying the DE:
yp'=a, yp"=0, so 2yp"+yp'-yp=a-ax-b=x+1, so a-b=1 and a=-1, so b=-2. Therefore, yp=-x-2
y=Ae½x+Be-x-x-2.
y'=½Ae½x-Be-x-1; y"=¼Ae½x+Be-x; y'(0)=0=A/2-B-1, A=2B+2, and y"(0)=1=A/4+B=(B+1)/2+B,
B+1+2B=2, B=⅓, A=8/3.
SOLUTION: y=(8/3)e½x+⅓e-x-x-2.
CHECK
y'=(4/3)e½x-⅓e-x-1, y"=⅔e½x+⅓e-x; y'(0)=4/3-1/3-1=0; y"(0)=2/3+1/3=1;
2y"+y'-y=(4/3)e½x+⅔e-x+(4/3)e½x-⅓e-x-1-(8/3)e½x-⅓e-x+x+2=x+1, so the solution checks out OK.